extreme f(x)=(x^4)/4-x^3+3

Lösung

extrempunkte

f (x) = \frac{x ^{4}}{4} - x^{3} + 3

Sattel (0, 3), Minimum (3, - \frac{15}{4})

Schritte zur Lösung

f^{'} (x) = x^{3} - 3 x^{2}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Absteigend

: 0 < x < 3,

Ansteigend

: 3 < x < \infty

Stecke

x = 0

\frac{x ^{4}}{4} - x^{3} + 3 : 3

Sattel (0, 3)

Stecke

x = 3

\frac{x ^{4}}{4} - x^{3} + 3 : - \frac{15}{4}

Minimum (3, - \frac{15}{4})

Sattel (0, 3), Minimum (3, - \frac{15}{4})

e x t re m e f (x) = - 3 (x + 2) e^{4 x}, (- 5, 0)

e x t re m e f (x) = \frac{x ^{4} - 4 x ^{3}}{2}

e x t re m e f (x) = 0.5 x^{2} + 15 x + 5000

e x t re m e f (x) = 3 x^{3} - 3 x^{2} - 3 x + 10

e x t re m e 0.5 x + 15 + \frac{5000}{x}